Differentiating a Stochastic Integral with a Moving Boundary

# Parametric Stochastic Integral


# Parametric Stochastic Integral












I need help.

Defining the parametric stochastic integral

$$ F_t = \int_t^T\xi(t,s)g(s)ds $$

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with $\xi$ a generic stochastic process such that $d\xi(t,s) = \mu(t,s)dt + \sigma(t,s)dW_t$, I'm trying to prove that

$\\\\$ $$ dF_t = - g(t)\xi(t,t)dt + \int_t^Td\xi(t,s)g(s)ds$$

My first attempt was as follows :

$$ \xi(t,s) = \xi(0,s) + \int_0^t\mu(u,s)du + \int_0^t \sigma(u,s)dW_u $$

and so

\begin{eqnarray*} F_t &=& \int_t^T\xi(0,s)g(s)ds + \int_t^T\int_0^t\mu(u,s)g(s)duds + \int_t^T\int_0^t\sigma(u,s)g(s)dW_uds\\ &=& \int_t^T\xi(s,s)g(s)ds - \int_t^T\int_t^s\mu(u,s)g(s)duds - \int_t^T\int_t^s\sigma(u,s)g(s)dW_uds \end{eqnarray*}

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Assuming suitable conditions to apply the stochastic Fubini theorem, we get

$\\\\$

\begin{eqnarray*} F_t = \int_t^T\xi(s,s)g(s)ds - \int_t^T\alpha(u,T)du - \int_t^T\beta(u,T)dW_u \end{eqnarray*}

with

\begin{eqnarray*} \alpha(u,T) = \int_u^T\mu(u,s)g(s)ds \quad \quad \text{and} \quad \quad \beta(u,T) = \int_u^T\sigma(u,s)g(s)ds \end{eqnarray*}

Applying Ito's lemma, we find

$\\\\$

\begin{eqnarray*} dF_t &=& -\xi(t,t)g(t)dt + \alpha(t,T)dt + \beta(t,T)dW_t\\ &=& -\xi(t,t)g(t)dt + \int_t^T\left(\mu(t,s)dt + \sigma(t,s)dW_t\right)g(s)ds\\ &=& -\xi(t,t)g(t)dt + \int_t^Td\xi(t,s)g(s)ds \end{eqnarray*}

Now, I have two questions :

- Is my proof correct ?

- Is there a more clever and faster answer ?

Thank you in advance for your answer.

## Answer by Kurt G. (score 1)

https://quant.stackexchange.com/a/67786

I am having trouble to understand your notation $$ \int_t^Td\xi(t,s)g(s)\,ds\,. $$ What is the meaning of this when you switch from the differential form $dF_t$ to the integral form $$ F_t=F_0-\int_0^tg(s)\,\xi(s,s)\,ds\,+\quad? $$ Surely, in the deterministic case when $\sigma\equiv 0\,$ we have by ordinary calculus $$ \frac{dF}{dt}=-\xi(t,t)\,g(t)+\int_t^T\frac{\partial}{\partial t}\xi(t,s)\,g(s)\,ds\,, $$ or, in integral form $$\tag{0} F_t=F_0-\int_0^tg(s)\,\xi(s,s)\,ds+\int_0^t\int_u^T\frac{\partial}{\partial u}\xi(u,s)\,g(s)\,ds\,du\,. $$ To get to the bottom of the stochastic case I consider only the case $\mu\equiv 0,\sigma\not\equiv 0,g\equiv1$ to simplify notation.

From $\xi(t,s)=\xi(0,s)+\int_0^t\sigma(u,s)\,dW_u$ we get (using stochastic Fubini) \begin{align} F_t&=\int_t^T\xi(t,s)\,ds=\int_t^T\xi(0,s)\,ds+\int_t^T\left(\int_0^t\sigma(u,s)\,dW_u\right)\,ds\\ &=\int_t^T\xi(0,s)\,ds+\int_0^t\int_t^T\sigma(u,s)\,ds\,dW_u\,. \end{align} By Ito's formula, $$\tag{1} dF_t=-\xi(0,t)\,dt+\left(\int_t^T\sigma(t,s)\,ds\right)dW_t-\left(\int_0^t\sigma(u,t)\,dW_u\right)\,dt\,. $$ The last term in (1) can be combined with the first term and gives \begin{align}\tag{2} dF_t&=-\xi(t,t)\,dt+\left(\int_t^T\sigma(t,s)\,ds\right)dW_t\,. \end{align} In integral form, (2) is $$\tag{3}\boxed{ F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^t\int_u^T\sigma(u,s)\,ds\,dW_u\,.} $$ By stochastic Fubini, this is $$\tag{4} F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^T\int_0^{s\wedge t}\sigma(u,s)\,dW_u\,ds\,. $$ Using $$ d\xi(t,s)=\sigma(t,s)\,dW_t $$ one could write (4) as $$\tag{5}\boxed{ F_t=F_0-\int_0^t\xi(s,s)\,ds+\int_0^T\big\{\xi(s\wedge t,s)-\xi(0,s)\big\} \,ds\,.} $$ It is fairly easy to see that (0) can also be written in the same form. In other words, (5) is the form that comprises the deterministic and the stochastic case.